When the string is cut, the only force producing torque about the hinge P is the weight of the rod ($Mg$). This force acts at the center of mass of the rod, which is at a distance of $L/2$ from the hinge P. Torque, $\tau = \text{Force} \times \text{perpendicular distance} = Mg \times \frac{L}{2}$ The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through one end is $I = \frac{ML^2}{3}$. According to Newton's second law for rotational motion, $\tau = I\alpha$, where $\alpha$ is the angular acceleration. Equating the two expressions for torque: $Mg \frac{L}{2} = \left(\frac{ML^2}{3}\right) \alpha$ $\alpha = \frac{Mg \frac{L}{2}}{\frac{ML^2}{3}} = \frac{3g}{2L}$